Linear Algebra Examples

$- 3 x + y - z = - 2$ , $- 3 x + z = 4$ , $y - 5 z = 0$

Write the system of equations in matrix form.

$[\begin{array}{c} - 3 & 1 & - 1 & - 2 \\ - 3 & 0 & 1 & 4 \\ 0 & 1 & - 5 & 0 \end{array}]$

Row reduce.

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Perform the row operation $R_{1} = - \frac{1}{3} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{1}{3} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} - \frac{1}{3} R_{1} & - \frac{1}{3} R_{1} & - \frac{1}{3} R_{1} & - \frac{1}{3} R_{1} \\ - 3 & 0 & 1 & 4 \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{1} = - \frac{1}{3} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{3} R_{1}$ .

$[\begin{array}{c} (- \frac{1}{3}) \cdot (- 3) & (- \frac{1}{3}) \cdot (1) & (- \frac{1}{3}) \cdot (- 1) & (- \frac{1}{3}) \cdot (- 2) \\ - 3 & 0 & 1 & 4 \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{1} = - \frac{1}{3} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ - 3 & 0 & 1 & 4 \\ 0 & 1 & - 5 & 0 \end{array}]$

Perform the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 3 \cdot R_{1} + R_{2} & 3 \cdot R_{1} + R_{2} & 3 \cdot R_{1} + R_{2} & 3 \cdot R_{1} + R_{2} \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{2} = 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ (3) \cdot (1) - 3 & (3) \cdot (- \frac{1}{3}) + 0 & (3) \cdot (\frac{1}{3}) + 1 & (3) \cdot (\frac{2}{3}) + 4 \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{2} = 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & - 1 & 2 & 6 \\ 0 & 1 & - 5 & 0 \end{array}]$

Perform the row operation $R_{2} = - 1 \cdot R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 1 \cdot R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ - 1 \cdot R_{2} & - 1 \cdot R_{2} & - 1 \cdot R_{2} & - 1 \cdot R_{2} \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{2} = - 1 \cdot R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 1 \cdot R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ (- 1) \cdot (0) & (- 1) \cdot (- 1) & (- 1) \cdot (2) & (- 1) \cdot (6) \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{2} = - 1 \cdot R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{1}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 1 & - 5 & 0 \end{array}]$

Perform the row operation $R_{1} = \frac{1}{3} R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{3} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{1}{3} R_{2} + R_{1} & \frac{1}{3} R_{2} + R_{1} & \frac{1}{3} R_{2} + R_{1} & \frac{1}{3} R_{2} + R_{1} \\ 0 & 1 & - 2 & - 6 \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{1} = \frac{1}{3} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{3} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{1}{3}) \cdot (0) + 1 & (\frac{1}{3}) \cdot (1) - \frac{1}{3} & (\frac{1}{3}) \cdot (- 2) + \frac{1}{3} & (\frac{1}{3}) \cdot (- 6) + \frac{2}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 1 & - 5 & 0 \end{array}]$

$R_{1} = \frac{1}{3} R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 1 & - 5 & 0 \end{array}]$

Perform the row operation $R_{3} = - 1 \cdot R_{2} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - 1 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ - 1 \cdot R_{2} + R_{3} & - 1 \cdot R_{2} + R_{3} & - 1 \cdot R_{2} + R_{3} & - 1 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 1 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 1 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ (- 1) \cdot (0) + 0 & (- 1) \cdot (1) + 1 & (- 1) \cdot (- 2) - 5 & (- 1) \cdot (- 6) + 0 \end{array}]$

$R_{3} = - 1 \cdot R_{2} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 0 & - 3 & 6 \end{array}]$

Perform the row operation $R_{3} = - \frac{1}{3} R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - \frac{1}{3} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ - \frac{1}{3} R_{3} & - \frac{1}{3} R_{3} & - \frac{1}{3} R_{3} & - \frac{1}{3} R_{3} \end{array}]$

$R_{3} = - \frac{1}{3} R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - \frac{1}{3} R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ (- \frac{1}{3}) \cdot (0) & (- \frac{1}{3}) \cdot (0) & (- \frac{1}{3}) \cdot (- 3) & (- \frac{1}{3}) \cdot (6) \end{array}]$

$R_{3} = - \frac{1}{3} R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & - \frac{1}{3} & - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 0 & 1 & - 2 \end{array}]$

Perform the row operation $R_{1} = \frac{1}{3} R_{3} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{3} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{1}{3} R_{3} + R_{1} & \frac{1}{3} R_{3} + R_{1} & \frac{1}{3} R_{3} + R_{1} & \frac{1}{3} R_{3} + R_{1} \\ 0 & 1 & - 2 & - 6 \\ 0 & 0 & 1 & - 2 \end{array}]$

$R_{1} = \frac{1}{3} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{3} R_{3} + R_{1}$ .

$[\begin{array}{c} (\frac{1}{3}) \cdot (0) + 1 & (\frac{1}{3}) \cdot (0) + 0 & (\frac{1}{3}) \cdot (1) - \frac{1}{3} & (\frac{1}{3}) \cdot (- 2) - \frac{4}{3} \\ 0 & 1 & - 2 & - 6 \\ 0 & 0 & 1 & - 2 \end{array}]$

$R_{1} = \frac{1}{3} R_{3} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & 0 & - 2 \\ 0 & 1 & - 2 & - 6 \\ 0 & 0 & 1 & - 2 \end{array}]$

Perform the row operation $R_{2} = 2 \cdot R_{3} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = 2 \cdot R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & - 2 \\ 2 \cdot R_{3} + R_{2} & 2 \cdot R_{3} + R_{2} & 2 \cdot R_{3} + R_{2} & 2 \cdot R_{3} + R_{2} \\ 0 & 0 & 1 & - 2 \end{array}]$

$R_{2} = 2 \cdot R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 2 \cdot R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 2 \\ (2) \cdot (0) + 0 & (2) \cdot (0) + 1 & (2) \cdot (1) - 2 & (2) \cdot (- 2) - 6 \\ 0 & 0 & 1 & - 2 \end{array}]$

$R_{2} = 2 \cdot R_{3} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & 0 & 0 & - 2 \\ 0 & 1 & 0 & - 10 \\ 0 & 0 & 1 & - 2 \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$x = - 2$

$y = - 10$

$z = - 2$

The solution is the set of ordered pairs that makes the system true.

$(- 2, - 10, - 2)$

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